= Solution
Put $s=1/n$, so $0<s<1$. For the <complex exponentiation> $z^s$, choose $-\pi/2<\arg z<3\pi/2$: the <branch cut> is the negative imaginary axis. Integrate $z^s/(a^2+z^2)$ around the upper half-plane semicircle, with a small semicircular indentation excluding the origin. The only enclosed <simple pole> is $z=ia$, with <residue>
$$
\operatorname{Res}_{z=ia}\frac{z^s}{a^2+z^2}=\frac{(ia)^s}{2ia}.
$$
The large arc is $O(R^{s-1})$ and the small arc is $O(\epsilon^{s+1})$, so both vanish. On the negative real segment, $z^s=e^{i\pi s}|z|^s$; accounting for the direction of traversal, the two real segments therefore sum to $(1+e^{i\pi s})I$, where $I$ is the desired positive-axis integral. The <residue theorem> gives
$$
(1+e^{i\pi s})I=\pi a^{s-1}e^{i\pi s/2}.
$$
Since $1+e^{i\pi s}=2e^{i\pi s/2}\cos(\pi s/2)$,
$$
\boxed{I=\frac{\pi a^{1/n-1}}{2\cos[\pi/(2n)]}.}
$$
The endpoint estimates also prove convergence. More generally the same calculation is the <fractional-power integral from an upper-half-plane contour>, valid for $-1<s<1$.
Back to article page
